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My goal is to compute

\begin{equation} I = \det \left(\mathbf{I} + \mathbf{A}\right) \end{equation} where $\mathbf{A}$ is a $n \times n$ checkerboard matrix filled with Catalan numbers: $$ \left\{ \begin{array}{crc} \mathbf{A}_{ij} = C_{p-1} \alpha^{2(p-1)} &\mbox{ if }& i+j=2p \mbox{, and with }C_p= \frac{1}{p+1}\binom {2p} {p}\\ \mathbf{A}_{ij} = 0 &\mbox{ if }& i+j \mbox{ is odd.} \end{array} \right. $$ with $\alpha >0$ a parameter.

Numerically, it seems that $I$ has a limit when $n\to +\infty$ if $\alpha<1/2$ and diverges to $\infty$ otherwise.

Any idea ?

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  • $\begingroup$ What are the first few values? Anything in the Online Encyclopedia of Integer Sequences? $\endgroup$ – Douglas Zare Apr 21 '15 at 12:54
  • $\begingroup$ what relation does $p$ have to $n$? $\endgroup$ – Suvrit Apr 21 '15 at 13:16
  • $\begingroup$ @Suvrit :p is just (i+j)/2 when i+j is even $\endgroup$ – user16215 Apr 21 '15 at 13:51
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    $\begingroup$ Maybe some ideas in cs.ucsb.edu/~omer/DOWNLOADABLE/catalan_hankel09.pdf may be helpful. or also in: arxiv.org/pdf/0709.3044.pdf $\endgroup$ – Suvrit Apr 21 '15 at 14:30
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    $\begingroup$ Note that there is no interaction between the even and odd coordinates in your matrix. In other words, if you rearrange rows and columns to put the even-numbered rows first and then the odd-numbered ones, you get a block-diagonal matrix and you should probably be looking at the blocks individually. It's also unclear at what numbers your matrix starts, you should write down the top left corner explicitly. There's a nice combinatorial interpretation for these; see jacobi.math.wvu.edu/~jerzy/research/21catalan.pdf. $\endgroup$ – Anton Malyshev Apr 21 '15 at 14:44
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If you consider the operator on functions that sends $f(x)$ to

$$Tf(y) = \int_{-2}^2 \frac{ \sqrt{4-x^2} }{2\pi } \frac{1}{ 1- \alpha x y} f(\alpha x) dx$$

then this is a well-defined integral operator from functions on $[-2\alpha,2\alpha]$ to functions on $[-2\alpha, 2\alpha]$ as long as $\alpha<1/2$.

Moreover, if $f(x)=x^{i-1}$, then the coefficient of $y^{j-1}$ in $Tf(y)$ is exactly $\mathbf A_{ij}$. So the limit of your determinant should be exactly the Fredholm determinant of this integral operator.

I'm not sure if this helps at all.

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Since $$\binom{2p}{p} =O\left(\frac{4^p}{\sqrt{p}}\right),$$ the coefficients of your matrix are unbounded when $\alpha > 1/2,$ while for $\alpha < 1/2$ the matrix $A$ is well-approximated by its piece where $i+j$ is small, so your observations are not surprising. The interesting case is $\alpha = 1/2,$ where I have no intuition.

EDIT A mathematica experiment seems to indicate that the determinants converge slowly for $\alpha = 1/2.$

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  • $\begingroup$ Thank you! Just a minor comment : the fact that the coefficient are unbounded does not imply that the determinant will be large, I think. $\endgroup$ – user16215 Apr 21 '15 at 19:35
  • $\begingroup$ @user16215 It does not logically imply it, it suggests it - notice that I never claimed that this was proof. $\endgroup$ – Igor Rivin Apr 21 '15 at 19:41

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